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The question is:

A uniform rod of length L stands vertically upright on a smooth floor in a position of unstable equilibrium. The rod is then given a small displacement at the top and tips over. What is the rod's angular velocity when it makes an angle of 30 degrees with the floor, assuming the rod does not slip?

I do know what steps to take to solve this problem, and I got the correct answer: $$\omega = {\left(\frac{24g}{13L}\right)}^{1/2}$$

The method is to apply the law of conservation of energy, and use a constraint relation of $v = \frac{\sqrt 3}4 L\omega$, to obtain the given answer. You can ask me if you need my help in getting the constraint relation.

My question is, why are we applying the law of conservation of energy? It is applied when only conservative forces act on a system, right? But here, along with the gravitational force of the earth, normal force by the floor is being applied too.

My other question is, whether the acceleration of the centre of mass of the rod is constant or not, which would mean that the normal force is constant. Thus, if that is so, we can find the angular velocity of the rod using the SUVAT equations.

That is it, I guess. Thank you for going through my question. Please feel free to comment on it.

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The normal force never does work (never positively or negatively affects the kinetic energy of an object). Work is done when a force is applied in the direction an object is moving (when $F\cdot v\neq0$). So the normal force could only ever do work if an object was somehow touching the floor but also moving away from the floor... which just doesn't make any sense and isn't possible. So in some sense, the normal force can be called a conservative force. Actually, so is static friction, which is also at play here right? Because the bottom of the rod doesn't slip along the ground. Again, a force is being applied from static friction, but only to an object that isn't moving so $F\cdot v=0$ because $v=0$. If the bar slips and experienced kinetic friction, it would lose energy - because kinetic friction is a force parallel to the ground acting on an object that is also moving parallel to the ground.

Indeed all the gravitational potential energy that this thing loses will be turned into kinetic energy.

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  • $\begingroup$ Sorry, I forgot to mention that the floor was smooth but I understood the friction part. What I did not understand was how the work done by the normal reaction is zero. Can you reply to my other question? Thanks. $\endgroup$
    – Aditya Bansal
    Commented Mar 23, 2023 at 11:22
  • $\begingroup$ The acceleration of the center of mass is not constant. In fact it's very small at first when the force of gravity is nearly cancelled out by the normal force. That's why it's a lot more convenient to use conservation of energy here. $\endgroup$
    – AXensen
    Commented Mar 23, 2023 at 11:26
  • $\begingroup$ Could you prove it though? Also, why is the normal reaction not doing any work here? I mean, If I am not wrong, in this case, the normal reaction points in the upward direction, and the centre of mass moves in a downward direction, so the work done is negative. $\endgroup$
    – Aditya Bansal
    Commented Mar 23, 2023 at 11:52
  • $\begingroup$ Easiest way to see it is to think of it as a body rotating around the point on the bottom. The normal force is applied at the center of rotation so it doesn't do any torque, and doesn't do any work on the rotational motion. But it seems you want me to frame it in terms of the center of mass plus rotation around the center point - in which case it does negative work on the center of mass motion but positive work on the rotational motion - those are equal and opposite and they cancel out. They cancel because the point on the object you apply a force to needs to be moving, not the center of mass. $\endgroup$
    – AXensen
    Commented Mar 23, 2023 at 11:57
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    $\begingroup$ Oooh, I see. I did not know that the point at which the force was being applied had to have been moving for any work to occur. Thank you for letting me know. Also, could you explain why the acceleration of the centre of mass is not constant? $\endgroup$
    – Aditya Bansal
    Commented Mar 23, 2023 at 12:24

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